# Tag Info

3

Let $q = p/(p-1)$. Then $p,q$ satisfy the requirements of Youngs inequality, so that $$|u|^{p-1} |u_{x_n}| \leq C [|u|^{q(p-1)} + |u_{x_n}|^p]= C [|u|^p + |u_{x_n}|^p].$$ The other terms (the signum and terms involving $\zeta$ can be bounded by constants).

2

Integration in polar coordinates (or spherical, in higher dimensions). Write $z=x+\rho\theta$ where $\theta\in S^{n-1}$ is a unit vector and $\rho\in [0,r]$. Then for any integrable $f$ we have $$\int_{B(x,r)} f(z)\,dz = \int_0^r \int_{S^{n-1}} f(x+\rho \theta) \rho^{n-1}\,d\theta \,d\rho \tag{1}$$ (Compare with $n=2$ case, when this is the usual polar ...

2

If $p\gt 1$, we define the $L^{p,\infty}$ semi-norm by $$\lVert f\rVert_{p,\infty}^p:=\sup_{t\gt 0}t^p\lambda\{s, |f(s)|\gt t\}$$ (this is equivalent to a norm, namely, $\sup_{A,\lambda(A)\in (0,\infty)}\mu(A)^{1/p-1}\int_A|f|\mathrm d\lambda$). If we define $x:=k^{1/m}$ and if we use the inequality, we obtain $$x^{2m\frac{N+1}N}\lambda\{|u|\lt ... 2 Hint: g=f_x so f_{xx}=g_x. Now h=g_x so f_{xxx}=?. Just add more dependent variables (unknown functions) 1 In various branches of mathematics it turns out to be profitable to shift attention from what some objects are to what they do. In case of functions, we shift attention from (locally integrable) functions f being some set of ordered pairs, etc to what it does: it defines a linear functional on the set of test functions, by$$\varphi \mapsto \int f\varphi ...

1

We want to estimate $\int_B |\nabla\bar u|^p$ in terms of $\int_{B^+} |\nabla u|^p$. To begin with, $$\int_B |\nabla \bar u|^p = \int_{B^+} |\nabla u|^p +\int_{B^-} |\nabla \bar u|^p$$ On $B^-$, the function $\bar u$ is the sum of two functions: $$v(x) = -3u(x_1,\ldots,x_{n-1},-x_n)$$ and $$w(x) = 4u(x_1,\ldots,x_{n-1},-\frac{x_n}{2})$$ Compute the ...

1

No, you can't have $p<1$ there. Take the constant function $u\equiv \lambda$. Your inequality becomes $$\|\lambda\|_{2^*}\le \|\lambda\|_1^p$$ which (if $p< 1$) fails when $\lambda$ is large enough. When you imagine an inequality you'd like to be valid, consider how it scales when $u$ is multiply by a positive number, or (when working on vector ...

1

If you solve the differential equation $\frac{d^2 X}{dx^2} - k^2 X = 0$, then let $X=e^{mx}$ so that you get the auxiliary equation $m^2-k^2=0$. That auxiliary equation has roots $m_1=k,m_2=-k$, which are real. If you plug the roots to the formula $$y=A e^{m_1 x}+ B e^{m_2 x}$$ you will get a solution with exponents. If you solve the differential equation ...

1

I'm assuming that instead of $3 x$ twice, one of them should be $3 y$. In this case, you can use superposition to solve separately the boundary value problem with the linear terms, and then the trigonometric terms, and then add together.. For example, $\psi = 3 x y$ solves laplaces equations with the boundary conditions $3 x$ and $3 y$. For the ...

1

Let's check: $$u_t = \frac{7}{64} u$$ $$\Delta u = -\frac{1}{8}u$$ hence $$\Delta\Delta u = \frac{1}{64}u$$ Yes, the sum is $0$. Yes, the solution increases in amplitude with time. Without the biLaplacian you would have time-reversed diffusion equation, which typically leads to concentration $u$ blowing up.

1

When a function $f$ is defined on $\mathbb R^n$, its Fourier transform $\hat f$ is also defined on $\mathbb R^n$. But it would be counterproductive to think of both $f$ and $\hat f$ as coinhabiting the same space. Forming expressions like $f+\hat f$ would be nonsensical. The functions are really defined on different spaces. To emphasize this distinction, ...

1

Yes. Even if $k$ is not smooth, we still have $u\equiv 0$ in $\Omega$. The reason is, we can extend the function $u$ to be zero outside $\Omega$. Since $u=\partial_\nu u=0$ on $\partial\Omega$, after this extension $u$ becomes a solution of the PDE in the whole space ${\mathbb R}^d$. The unique continuation theorem says that: if the solution $u$ vanishes ...

1

Weak convergence in $L^2(0,T;W^{1,2}(\Omega))$ means by definition that we have $u^*(u_k) \to u^*(u)$ for any functional $u^*\in L^2(0,T; W^{1,2}(\Omega))^* = L^2(0,T;W^{1,2}(\Omega)^*)$. To see that this implies weak convergence in $L^2(0,T;L^2(\Omega))$, note that $W^{1,2}(\Omega) \to L^2(\Omega)$ is a continuous injection. Hence, its dual, the ...

1

Hint: Your equation can be written $$\partial_x (\log u-ca) + i\partial_y(\log u - ca) = 0$$ This equation has the solution $\log u - ca = g$ for any $g$ that satisfy $$\partial_x g + i\partial_yg = 0$$

1

If you change your variables using a Lie group, you can find the most general solution of this pde. Use $$G(x,y,z)=(\lambda x,\lambda^\beta y,\lambda^\alpha z)\lambda_o=1$$ such that $x'=\lambda x$, $y'=\lambda^\beta y$, $z'=\lambda^\alpha z$, then $$\frac{\partial z'}{\partial x'}+\frac{\partial z'}{\partial y'}=\frac{\lambda^\alpha \partial z}{\lambda ... 1 Suppose the parameters for characteristics are \eta and \xi. The characteristics are defined by$$ \frac{\partial t}{\partial \eta} = 1, \frac{\partial x}{\partial \eta} = u, \frac{\partial u}{\partial \eta} = 0.  Solving gives $t(\eta, \xi) = \eta + c_1(\xi)$, $u(\eta, \xi) = c_3(\xi)$, $x(\eta, \xi) = c_3(\xi)\eta + c_2(\xi)$. Pick the ...

1

This is not true. Indeed, $f\in H^1(R^{n+1})\Leftrightarrow f$ and all of its first derivatives (in the sense of distributions) are in $L^2$. But if $f\in H^1(R,H^1(R^n))$, then necessarily $\frac{\partial^2 f}{\partial t\partial x_j}$ also lie in $L^2$, which is not the case for a general function in $H^1(R^{n+1})$.

Only top voted, non community-wiki answers of a minimum length are eligible